\[ \int_{0}^{1}{\frac{\arctan{\sqrt{x^{2}+2}}}{(x^{2}+1)\sqrt{x^{2}+2}}dx} \]

(Ahmed's integral)

Solution:

\[ \frac{\pi^{2}}{16}=\int_{0}^{1}{\int_{0}^{1}{\frac{dxdy}{(1+x^2)(1+y^2)}}} \]
\[=\int_{0}^{1}{\int_{0}^{1}{\frac{1}{(1+x^{2})(2+x^2+y^2)}+\frac{1}{(1+y^{2})(2+x^2+y^2)}dxdy}}\]
\[=2\int_{0}^{1}{\int_{0}^{1}{\frac{1}{(1+x^2)(2+x^2+y^2)}dydx}} \]
\[=2\int_{0}^{1}{\frac{1}{(1+x^2)\sqrt{2+x^2}}\arctan{\frac{1}{\sqrt{2+x^2}}}dx} \]
\[ =2\int_{0}^{1}{\left(\frac{\pi}{2(1+x^2)\sqrt{2+x^2}}-\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}} \right)dx} \]
\[ =\frac{\pi^{2}}{6}-2\int_{0}^{1}{\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}}dx} \]
\[ \Rightarrow \int_{0}^{1}{\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}}dx}=\frac{5}{96}\pi^{2}\]